Comparison Test Comparison Test

Direct Comparison Test

If

\sum a_n
and
\sum b_n
are positive series, where
a_n \leq b_n
for all
n
, then:
If
\sum b_n
converges, then
\sum a_n
also converges.
If
\sum a_n
diverges, then
\sum b_n
also diverges.

In other words:
If a series is always smaller than a converging series, then it must also converge.
If a series is always larger than a diverging series, then it must also diverge.

This test does not need to pass for all the terms. It just needs to pass for all terms beyond a certain point, as the sum of finitely many finite terms is finite. In other words,

\sum_{n=1}^{N} a_n
is finite if
N
and
a_n
are finite.

Limit Comparison Test

The Limit Comparison Test states that if

\sum a_n
and
\sum b_n
are series with positive terms, then:

If
\lim_{n \to \infty} a_n/b_n = c
converges, then
\sum a_n
and
\sum b_n
either both converge or both diverge.

Examples

Example 1: Convergence using Direct Comparison Test
Determine whether the series

\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}
converges or diverges.
\begin{align*} \text{For all } n &\geq 1, \\ \quad n^2 + 1 &> n^2 \\ \frac{1}{n^2 + 1} &< \frac{1}{n^2} \end{align*}
We know that
\sum_{n=1}^{\infty} \frac{1}{n^2}
is a p-series with
p = 2 > 1
, so it converges.
Since
0 < \frac{1}{n^2 + 1} < \frac{1}{n^2}
for all
n \geq 1
, by the Direct Comparison Test, the series
\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}
also converges.

Example 2: Divergence using Direct Comparison Test
Determine whether the series
\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} - 0.5}
converges or diverges.
\begin{align*} \text{For } n &\geq 1, \\ \sqrt{n} &> \sqrt{n} - 0.5 \\ \frac{1}{\sqrt{n}} &< \frac{1}{\sqrt{n} - 0.5} \end{align*}
We know that
\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{n^{1/2}}
is a p-series with
p = \frac{1}{2} < 1
, so it diverges.
Since
\frac{1}{\sqrt{n} - 0.5} > \frac{1}{\sqrt{n}} > 0
for all
n \geq 1
, by the Direct Comparison Test, the series
\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} - 0.5}
also diverges.

Example 3: Convergence using Limit Comparison Test
Determine whether the series
\sum_{n=1}^{\infty} \frac{3n - 2}{n^3 + 4n + 1}
converges or diverges.
\begin{align*} \text{Let } a_n &= \frac{3n - 2}{n^3 + 4n + 1} \\ \text{Let } b_n &= \frac{1}{n^2} \end{align*}
Consider the limit:
\begin{align*} \lim_{n \to \infty} \frac{a_n}{b_n} &= \lim_{n \to \infty} \frac{\frac{3n - 2}{n^3 + 4n + 1}}{\frac{1}{n^2}} \\ &= \lim_{n \to \infty} \frac{n^2(3n - 2)}{n^3 + 4n + 1} \\ &= \lim_{n \to \infty} \frac{3n^3 - 2n^2}{n^3 + 4n + 1} \\ &= \lim_{n \to \infty} \frac{3 - \frac{2}{n}}{1 + \frac{4}{n^2} + \frac{1}{n^3}} = 3 \end{align*}
Since
\lim_{n \to \infty} \frac{a_n}{b_n} = 3
is a finite positive number, and
\sum_{n=1}^{\infty} \frac{1}{n^2}
converges (p-series with
p = 2 > 1
), by the Limit Comparison Test, the series
\sum_{n=1}^{\infty} \frac{3n - 2}{n^3 + 4n + 1}
also converges.